Answer : The value of rate constant at temperature 119 K is 2.46 E-29
Explanation :
As we are the rate law expression as:
![Rate=1.4\times 10^{-2}[NO_2]^2](https://tex.z-dn.net/?f=Rate%3D1.4%5Ctimes%2010%5E%7B-2%7D%5BNO_2%5D%5E2)
..........(1)
The general rate law expression will be:
![Rate=k[NO_2]^2](https://tex.z-dn.net/?f=Rate%3Dk%5BNO_2%5D%5E2)
............(2)
By comparing equation 1 and 2 we get:
Now we have to calculate the rate constant at temperature 119 K.
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= ?
= activation energy for the reaction = 80.0 kJ/mole = 80000 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature = 500 K
= final temperature = 119 K
Now put all the given values in this formula, we get:
![\log (\frac{K_2}{1.4\times 10^{-2}})=\frac{80000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{500}-\frac{1}{119}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7B1.4%5Ctimes%2010%5E%7B-2%7D%7D%29%3D%5Cfrac%7B80000J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B500%7D-%5Cfrac%7B1%7D%7B119%7D%5D)
Therefore, the value of rate constant at temperature 119 K is 2.46 E-29
Answer is: A) 7.84 g.
V(Mg(NO₃)₂) = 151 mL ÷ 1000 mL/L.
V(Mg(NO₃)₂) = 0.151 L; volume of the magnesium nitrate.
c(Mg(NO₃)₂) = 0.352 M; molarity of the solution.
n(Mg(NO₃)₂) = V(Mg(NO₃)₂) · c(Mg(NO₃)₂).
n(Mg(NO₃)₂) ) = 0.151 L · 0.352 mol/L.
n(Mg(NO₃)₂) = 0.0531 mol; amount of the substance.
M(Mg(NO₃)₂) = Ar(Mg) + 2Ar(N) + 6Ar(O) · g/mol.
M(Mg(NO₃)₂) = 24.3 + 2·14 + 6·16 · g/mol.
M(Mg(NO₃)₂) = 148.3 g/mol; molar mass.
m(Mg(NO₃)₂) = n(Mg(NO₃)₂) · M(Mg(NO₃)₂).
m(Mg(NO₃)₂) = 0.0531 mol · 148.3 g/mol.
m(Mg(NO₃)₂) = 7.84; mass of magnesium nitrate.
Find the mass of the empirical formula.
You must be given a sample of some kind to calculate the weight or know how many moles are present. Then you figure out what one mol would be. The key step is multiplying the empirical formula numbers by what it takes to make 1 mol.
It would be clearer if we were working from some choices.
Answer: 4.56 Joules
Explanation:-
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
Q = Heat absorbed = ?
m= mass of substance = 2.5 g
c = specific heat capacity = 
Initial temperature = 
= 25.0°C
Final temperature = 
= 29.9°C
Change in temperature ,
Putting in the values, we get:
The heat absorbed by gallium must be 4.56 Joules
